A) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{7}\,\frac{h}{2\pi },\,\sqrt{2}\,\frac{h}{2\pi }\]
B) \[0\,\frac{\sqrt{6}h}{2\pi },\,\frac{\sqrt{2}h}{2\pi }\]
C) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{6}\frac{6}{2\pi },\,\sqrt{2}\frac{h}{2\pi }\]
D) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{6}\frac{h}{2\pi },\,\sqrt{3}\frac{h}{2\pi }\]
Correct Answer: C
Solution :
[c] Forelectron\[L=\sqrt{2(2+1)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }\] |
Forelectron\[L=\sqrt{1(1+1)}\frac{h}{2\pi }=\sqrt{2}.\frac{h}{2\pi }\] |
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